GMAT Mixtures Guide with Three Fully-Worked OG Examples

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Hi GMATters,

Here is my Mixtures Question guide, with three fully-worked OG examples.

Mixtures questions are really easy if you know how to lay them out, and here I explain my method to do so.

I've provided an excerpt below, but you can find the entire thing at ... uestions/

The Ultimate Guide to GMAT Mixtures Questions

The beauty of Mixtures questions is that they are relatively simple once we get a couple of simple concepts straight.

One nice thing about Mixture questions is that they are best learned through example, so let’s jump right in.

Let’s take a hypothetical case where we have a 5-liter bucket that is full of a mixture that is a% Coca-Cola and a 2-liter bucket that is full of a mixture that is b% Coca-Cola. You can imagine that the rest of it is whatever you like.
First question: how much liquid do we have in total?

That is as simple as it seems: just the volumes of the full buckets added together:

5 liters + 2 liters = 7 liters

Now, let’s say that if we add the two buckets together, we’ll get a mixture that is c% Coca-Cola. Now we have as much as we need for a standard GMAT Mixtures setup:

(a/100)(5 liters)+(b/100)(2 liters) = (c/100)(7 liters)

And that’s really it. In a live question, you’d see it in a form where you’re given two variables and asked to find the third. That is, for example, you’re told a and c and have to work out b.

It’s also possible that you can use the same information to collapse one variable to find the ratio between two of the others. We’ll look at cases that do both of these things.

First, let’s establish the nice big “General Mixture Equation.” In essence, your mixture should look something like this, but this is more of a roadmap than a formula–sort of like putting linear equations into y=mx+b or quadratics into ax^2+bx + c =0.
General GMAT Mixtures Equation

(a/100)(X)+(b/100)(Y) = (c/100)(X+Y)

Now let’s move on to some worked Official examples.
Chapter 1: A Classic Mixtures Question

The classic “difficult” mixture question from the OG is this one. We don’t need to worry about answer choices here because mixture questions, should, in principle, be objectively solvable.

Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of the mixture is X ?

In this case, it’s useful to write out all the details. As normal, put the percents into fractions as soon as possible:

X: 2/5 ryegrass**, 3/5** bluegrass

Y: 1/4 ryegrass**, 3/4** fescue

Total (X+Y): 3/10 ryegrass

What we’re looking for: X/(X+Y) as a percent (times 100)

t’s probably obvious by this point that we don’t need to worry about anything except the ryegrass. That is, ignoring everything else, we find this:

(2/5)X+(1/4)Y = 3/10(X+Y)


Read the rest here: ... questions/
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